标签：10 Code 题目 记录 处理 2021.12 DAG 处理器 气球
CF909E Coprocessor
题目大意
给定一张 DAG，其中含 n 个点和 m 条有向边。
给定两种处理器，主处理器和副处理器以及对应的点（主处理器一次只可处理一个点，副处理器一次可处理多个点）
求副处理器最少运行次数
思路
因为是 DAG ，所以想到拓扑排序。
题目要求副处理器运行次数最少，相当于要主处理器运行次数尽可能多。所以我们可以先把能处理的主处理器的任务一次性处理完，然后再处理副处理器的。
实现中，开两个队列维护两个处理器就OK了。
Code
<iframe sandbox="allow-scripts allow-same-origin" src="https://carbon.vercel.app/embed?bg=rgba%28255%2C255%2C255%2C0%29&t=one-dark&wt=none&l=text%2Fx-c%2B%2Bsrc&ds=true&dsyoff=0px&dsblur=26px&wc=true&wa=true&pv=37px&ph=38px&ln=true&fl=1&fm=Fira+Code&fs=14.5px&lh=116%25&si=false&es=4x&wm=false&code=%252F%252F%2520Problem%253A%2520CF909E%2520Coprocessor%250A%252F%252F%2520URL%253A%2520https%253A%252F%252Fwww.luogu.com.cn%252Fproblem%252FCF909E%250A%252F%252F%2520Memory%2520Limit%253A%2520250%2520MB%250A%252F%252F%2520Time%2520Limit%253A%25201500%2520ms%250A%252F%252F%2520Author%253AarlenWKX%250A%250A%2523%2520include%2520%253Cbits%252Fstdc%252B%252B.h%253E%250A%250Ausing%2520namespace%2520std%253B%250A%250Aconst%2520int%2520N%2520%253D%25201e5%2520%252B%25205%252C%2520M%2520%2520%253D%25201e5%2520%252B%25205%253B%250A%250Aint%2520n%252C%2520m%252C%2520a%2520%255B%2520N%2520%255D%253B%250Aint%2520had%2520%255B%2520N%2520%255D%252C%2520nxt%2520%255B%2520M%2520%255D%252C%2520vct%2520%255B%2520M%2520%255D%252C%2520edgenum%253B%250Aint%2520ind%2520%255B%2520N%2520%255D%253B%250Aqueue%2520%253C%2520int%2520%253E%2520q%2520%255B%25202%2520%255D%253B%250Aint%2520tot%252C%2520ans%253B%250A%250Avoid%2520adddoge%2520%28%2520int%2520u%252C%2520int%2520v%2520%29%250A%257B%250A%2509%250A%2509edgenum%2520%252B%252B%253B%250A%2509%250A%2509nxt%2520%255B%2520edgenum%2520%255D%2520%253D%2520had%2520%255B%2520u%2520%255D%253B%250A%2509vct%2520%255B%2520edgenum%2520%255D%2520%253D%2520v%253B%250A%2509had%2520%255B%2520u%2520%255D%2520%253D%2520edgenum%253B%250A%2509%250A%257D%250A%250Aint%2520main%2520%28%2520%29%250A%257B%250A%2509%250A%2509scanf%2520%28%2520%2522%2525d%2525d%2522%252C%2520%2526%2520n%252C%2520%2526%2520m%2520%29%253B%250A%2509%250A%2509for%2520%28%2520int%2520i%2520%253D%25201%253B%2520i%2520%253C%253D%2520n%253B%2520i%2520%252B%252B%2520%29%2520scanf%2520%28%2520%2522%2525d%2522%252C%2520%2526%2520a%2520%255B%2520i%2520%255D%2520%29%253B%250A%2509%2509%250A%2509for%2520%28%2520int%2520i%2520%253D%25201%252C%2520x%252C%2520y%253B%2520i%2520%253C%253D%2520m%253B%2520i%2520%252B%252B%2520%29%250A%2509%257B%250A%2509%2509scanf%2520%28%2520%2522%2525d%2525d%2522%252C%2520%2526%2520x%252C%2520%2526%2520y%2520%29%253B%250A%2509%2509adddoge%2520%28%2520%252B%252B%2520y%252C%2520%252B%252B%2520x%2520%29%253B%250A%2509%2509ind%2520%255B%2520x%2520%255D%2520%252B%252B%253B%250A%2509%2509%250A%2509%257D%250A%2509%250A%2509for%2520%28%2520int%2520i%2520%253D%25201%253B%2520i%2520%253C%253D%2520n%253B%2520i%2520%252B%252B%2520%29%250A%2509%2509if%2520%28%2520%21%2520ind%2520%255B%2520i%2520%255D%2520%29%2520q%2520%255B%2520a%2520%255B%2520i%2520%255D%2520%255D.push%2520%28%2520i%2520%29%253B%250A%2509%250A%2509while%2520%28%2520tot%2520%253C%2520n%2520%29%250A%2509%257B%250A%2509%2509%250A%2509%2509if%2520%28%2520%21%2520q%2520%255B%25200%2520%255D.empty%2520%28%2520%29%2520%29%250A%2509%2509%257B%250A%2509%2509%2509%250A%2509%2509%2509while%2520%28%2520%21%2520q%2520%255B%25200%2520%255D.empty%2520%28%2520%29%2520%29%250A%2509%2509%2509%257B%250A%2509%2509%2509%2509%250A%2509%2509%2509%2509int%2520u%2520%253D%2520q%2520%255B%25200%2520%255D.front%2520%28%2520%29%253B%2520q%2520%255B%25200%2520%255D.pop%2520%28%2520%29%253B%250A%2509%2509%2509%2509tot%2520%252B%252B%253B%250A%2509%2509%2509%2509for%2520%28%2520int%2520e%2520%253D%2520had%2520%255B%2520u%2520%255D%253B%2520e%253B%2520e%2520%253D%2520nxt%2520%255B%2520e%2520%255D%2520%29%250A%2509%2509%2509%2509%257B%250A%2509%2509%2509%2509%2509int%2520v%2520%253D%2520vct%2520%255B%2520e%2520%255D%253B%2520ind%2520%255B%2520v%2520%255D%2520--%253B%250A%2509%2509%2509%2509%2509if%2520%28%2520%21%2520ind%2520%255B%2520v%2520%255D%2520%29%2520q%2520%255B%2520a%2520%255B%2520v%2520%255D%2520%255D.push%2520%28%2520v%2520%29%253B%250A%2509%2509%2509%2509%257D%250A%2509%2509%2509%2509%250A%2509%2509%2509%257D%250A%2509%2509%2509%250A%2509%2509%257D%250A%2509%2509else%250A%2509%2509%257B%250A%2509%2509%2509%250A%2509%2509%2509ans%2520%252B%252B%253B%250A%2509%2509%2509%250A%2509%2509%2509while%2520%28%2520%21%2520q%2520%255B%25201%2520%255D.empty%2520%28%2520%29%2520%29%250A%2509%2509%2509%257B%250A%2509%2509%2509%2509int%2520u%2520%253D%2520q%2520%255B%25201%2520%255D.front%2520%28%2520%29%253B%2520q%2520%255B%25201%2520%255D.pop%2520%28%2520%29%253B%250A%2509%2509%2509%2509tot%2520%252B%252B%253B%250A%2509%2509%2509%2509for%2520%28%2520int%2520e%2520%253D%2520had%2520%255B%2520u%2520%255D%253B%2520e%253B%2520e%2520%253D%2520nxt%2520%255B%2520e%2520%255D%2520%29%250A%2509%2509%2509%2509%257B%250A%2509%2509%2509%2509%2509int%2520v%2520%253D%2520vct%2520%255B%2520e%2520%255D%253B%2520ind%2520%255B%2520v%2520%255D%2520--%253B%250A%2509%2509%2509%2509%2509if%2520%28%2520%21%2520ind%2520%255B%2520v%2520%255D%2520%29%2520q%2520%255B%2520a%2520%255B%2520v%2520%255D%2520%255D.push%2520%28%2520v%2520%29%253B%250A%2509%2509%2509%2509%257D%250A%2509%2509%2509%2509%250A%2509%2509%2509%257D%250A%2509%2509%2509%250A%2509%2509%257D%250A%2509%25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style="width: 1000px; height: 1714px; border: 0; transform: scale(1); overflow: hidden"> </iframe>CF725D Contest Balloons
题目大意
ACM比赛，大家都知道。AC一题会有一个气球。 现在有n(2<=n<=300000)n(2<=n<=300000) 支队伍，每支队伍的重量是 wi ，拥有 ti 个气球(wi,ti<=10^18)，当一支队伍的气球个数比它的重量都要大时，这个队伍就会飘起来，从而被取消比赛资格。 现在你带领的是1号队，你希望你队伍的名次尽可能靠前，你是个有原则的人，不会偷气球，但你可以把气球送给别的队伍，让他们飞起来。 求最终你的队伍所获得的最好名次
思路
显然，这又㕛叒叕是个贪心。为了尽可能多的迫害帮助其他队伍，我们应该优先将自己的气球分给弱势群体离飘起来所需要的气球数小且气球数量比我们多的队伍，这用排序+扫描+优先队列可以实现。
某队伍：我可谢谢您了
Code
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Code
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